To determine which of 300 adults have a colorectal cancer that affects 2% of the population, a stool sample is taken from each individual. The 300 adults are a random sample from the population. Instead of testing all 300 stool samples, the samples are grouped into 30 groups of 10 each, stool from each of the 10 samples in each group are mixed, and each of the 30 mixtures is tested.
a) What is the probability that any such mixture will contain the stool of at least one affected person and hence test positive?
b) In the 30 mixtures of stool, what is the expected number of mixtures that will test positive?
This is what i have tried and i am unsure if they are correct...
a) Let X be the r.v. of the number of affected person.
X ~ Bin(10, 0.02)
P(X≥1) = 0.183
b) Let Y be the r.v. of the number of mixtures that will test positive.
E(Y) = 1x0.183 + 2x0.183 + 3x0.183 + ... + 30x0.183
Any help or guidance will be greatly appreciated, thank you!
Part (a) is OK.
Part (b) is not. To work out the expectation the way you look to be trying to you should have $$ E(Y)=0\times b(0;30,0.183)+1 \times b(1;30,0.183)+ \dots +30 \times b(30;30,0.183) $$ where $b(k;n,p)$ is the probability mass function for the binomial distribution with parameters $n$ and $p$. However you don't want to do it like that. The expected number of occurrences in $N$ trials of events that have a probability of occurring in a single trial of $p$ is $Np$. So in this case $E(Y)=30\times0.183=5.49$